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JEE Main Physics Class 11 Physical world and Measurements Part 1 Questions
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© examsnet.com
Question : 4
Total: 100
The pitch of the screw gauge is
1
m
m
and there are 100 divisions on the circular scale. When nothing is put in between the jaws, the zero of the circular scale lies 8 divisions below the reference line. When a wire is placed between the jaws, the first linear scale division is clearly visible while 72 nd division on circular scale coincides with the reference line. The radius of the wire is
[25 Feb 2021 Shift 1]
1.64
m
m
0.82
m
m
1.80
m
m
0.90
m
m
Validate
Solution:
Given, pitch of screw gauge,
P
=
1
m
m
Number of division,
n
=
100
∴
Least count
(
L
C
)
=
P
n
=
1
∕
100
=
0.01
m
m
As, zero of circular division lies 8 divisions below.
∴
Zero error
=
8
×
L
C
=
8
×
0.01
=
0.08
m
m
Since, 1st linear scale division coinside with 72nd circular scale division.
∴
R
a
d
i
u
s
(
r
)
=
[
P
+
(
72
×
L
Q
−
Zero error
]
2
=
[
1
+
(
72
×
0.01
)
−
0.08
]
∕
2
=
(
1.72
−
0.08
)
∕
2
=
1.64
∕
2
=
0.82
m
m
© examsnet.com
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